Proof. The case of is obvious; otherwise, this follows from as proved in Exercise 1.

Separability

Definition. Let G be a group. The profinite topology on G is the coarsest topology such that every homomorphism from G to a finite group (equipped with the discrete topology) is continuous.

Definition. A subgroup H is separable in G if H is closed in the profinite topology of G.

Exercise 3. Let G be a group. is separable if and only if for all , there exists a homomorphism to a finite group such that . Note that if X = {1}, this is equivalent to: G is RF if and only if {1} is separable.

Hint. For the “if” direction, let and consider . For the other direction, use the definition of subbase and that .

Lemma 3. Let G be a group. A subgroup H of G is separable if and only if for all , there exists a finite-index subgroup such that and .

Proof. In the “only if” direction, by the previous exercise, for all , there exists a homomorphism to a finite group such that . Then . Conversely, let . By hypothesis, there exists a finite-index subgroup such that and . Let . Note that this is a finite number of intersections (, to be precise). There exists a finite quotient . Then . Therefore, , i.e., , and the lemma follows by the previous exercise.

Scott’s Criterion (1978). Let X be a Hausdorff topological space and . Let be a covering and . Then H is separable in G if and only if for any compact , there exists and intermediate finite-sheeted cover such that embeds into .